Let $A,B \subset X$, where $X$ is a Hilbert space, be orthogonal subspaces.
Do we have
$$\overline{A \oplus B} = \overline{A} \oplus \overline{B}, \tag{1}$$
where $\oplus$ denotes the orthogonal direct sum.
Some background: When discussing pseudoinverses of linear operator on Hilbert spaces we defined the domain of the pseudoinverse $T^+$ of an operator $T: X \to Y$, as $D(T^+) = \text{ran(T)} \oplus \text{ran}(T)^{\perp}$.
A remark states that this domain is dense as $\overline{D(T^+)} = \overline{\text{ran}(T)} \oplus \text{ran(T)}^{\perp} = Y$.
I want to the reasoning behind this statement.
If $(1)$ would hold, the result would follow, as $\text{ran}(T)^{\perp}$ is already closed, i.e. $\overline{\text{ran}(T)^{\perp}} = \text{ran}(T)^{\perp}$.
If yes, can you please provide me a hint to proof it.
My thoughts:
"$\subset$": Let $(a_n + b_n)_{n \in \mathbb{N}} \subset A \oplus B$ with $(a_n + b_n)_n \to c \in \overline{A \oplus B}$.
Now we have to show that $c = a + b$ with $ a \in \overline{A}$ and $b \in \overline{B}$, right?
Best Answer
To answer your main highlighted question, (1) is not true for general topological vector spaces, as is shown by this answer.
What we do have though, for any Hilbert space $H$, is if $A,B\subset H$ are orthogonal, then $\bar A\oplus \bar B \subset \overline{A\oplus B}$. This is simple to prove. We first note that the continuity of the inner product means that if $A$ is orthogonal to $B$ then $\bar A$ is orthogonal to $\bar B$. Now take $x\in \bar A\oplus \bar B $, which by the above implies $x=a+b$ with $a\in \bar A$ and $b\in\bar B$. We have sequences $(a_n)\subset A$ converging to $a$ and $(b_n)\subset B$ converging to $b$. Clearly $x=\lim_{n\to\infty} a_n+b_n$ using the standard add and subtract trick in the norm, which finishes the proof, because $(a_n+b_n)\subset A\oplus B$.
This is enough for your problem, because as you note $Y=\overline{\operatorname{ran}(T)}\oplus \operatorname{ran}(T)^{\perp}=\overline{\operatorname{ran}(T)}\oplus \overline{\operatorname{ran}(T)^{\perp}}\subset\overline{D(T^+)}$.